Question:
\(\lim\limits_{t\rightarrow-3}\frac{x^2+16x+39}{2x^2+7x+3}\) is equal to
\(\lim\limits_{t\rightarrow-3}\frac{x^2+16x+39}{2x^2+7x+3}\) is equal to
Updated On: Jun 10, 2024
- 2
- \(\frac{8}{3}\)
- \(\frac{-8}{3}\)
- -2
- 0
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The Correct Option is D
Solution and Explanation
The correct option is (D): -2
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